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Let $G$ be a group such that its order is a product of distinct primes $p_1, \dots, p_n$ and let $P_i$ denote each Sylow $p_i$-subgroup. Is $P_1 \dots P_n$ (the internal or Frobenius product) equal to $G$, that is, $G = P_1 \dots P_n$?

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  • $\begingroup$ @ChrisGerig oh, okay! Is there a way to move the question or I should delete it here and re-post it there? $\endgroup$ Nov 4, 2020 at 0:13
  • $\begingroup$ @YCor Indeed, $G = P_1 \dots P_n$ is more precise. What notation would you suggest for the product instead of $P_1 \dots P_n$? I am happy with it, but I struggle to think of anything better. $\endgroup$ Nov 5, 2020 at 13:13
  • $\begingroup$ @YCor Indeed. Everything should be correct now. Thanks for the feedback! $\endgroup$ Nov 5, 2020 at 16:12

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For $G$ finite, one of the many theorems of P. Hall is that your condition holds whenever $G$ is solvable (regardless of order). A note of Rowley and Holt discusses the general problem (and includes a reference for said result of Hall), and provides a few non-solvable examples. They also show that such a product does not exist for the finite simple group $G=U_3(3)$. So in full generality the answer is "no". However, if $|G|$ is square-free, as you assume, then $G$ is solvable (see also these notes for more details), so the answer will be "yes" in this case.

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    $\begingroup$ In the original question $|G|$ is supposed to be a product of distinct primes. The example of Rowley and Holt is interesting and I am glad to have seen it, but their group has order $2^5.3^3.7$. $\endgroup$ Nov 4, 2020 at 0:42
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    $\begingroup$ I did not know that, though perhaps I should have done. The MSE question that you mention does not really give a self-contained proof, but I found one in Sections 5-6 of scholar.rose-hulman.edu/cgi/…. I think that this essentially qualifies as an answer to the OPs question. $\endgroup$ Nov 4, 2020 at 1:02
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There is a more direct proof than quoting the fairly deep Theorem of P. Hall, but you do need to know a little transfer theory. The argument that follows is well-known and may be found in many group theory texts. We proceed by induction, there being nothing to prove when $n = 1$. Suppose then that $n > 1$ and that the result is true for smaller values of $n$. If $|G| = p_{1}p_{2} \ldots p_{n}$ where $p_{1} < p_{2} < p_{3} < \ldots < p_{n}$ are primes, and if we let $P_{i}$ be a Sylow $p_{i}$-subgroup of $G$ for each $i$, then we note that the order of $N_{G}(P_{1})/C_{G}(P_{1})$ divides $p_{1}-1.$ But since $p_{1}$ is the smallest prime divisor of $|G|$, we see that $N_{G}(P_{1}) = C_{G}(P_{1})$.

By Burnside's transfer theorem, $G$ has a normal $p_{1}$-complement, which means that $G$ has a normal subgroup $H_{1}$ of order $p_{2}p_{3} \ldots p_{n}.$ Then $H_{1}$ contains all elements of $G$ of order coprime to $p_{1}$, and we have $G = H_{1}P_{1} = P_{1}H_{1}$, since $H_{1} \lhd G$.

By induction, we have $H_{1} = P_{2} P_{3} \ldots P_{n}$, so that $G = P_{1}H_{1} = P_{1}P_{2} \ldots P_{n}.$

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